# Thread: Difficult series section - Abel/Dirichlet's Test

1. ## Difficult series section - Abel/Dirichlet's Test

I'm having a lot of trouble with this section on series, as many of you who viewed and responded (thanks!) to my other posts know. I can't do about 20 of the problems provided, out of 30, so I think I'm missing something "integral" here.

I re-read the chapter, and I think that perhaps what I'm missing is the proper use of Dirichlet's and Abel's test. It seems, from my reading, that these two tests are really pretty much equivalent. I present Abel's test, as it is written in my book, below:

Let $\sum a_n$ be a covergent series of complex terms, and let $\{b_n\}$ be a monotonic convergent sequence of real terms. Then the series $\sum a_n b_n$ converges.

Here's a few example problems that I'm stuck on. I'm not presenting a few because I want/expect answers on how to solve all of them, but rather I am presenting these to see if Abel/Dirichlet's tests would even be appropriate for them:

$\sum_{n=1}^\infty \log \left(n \sin \frac 1 n \right)$
$\sum_{n=1}^\infty (-1)^n \left ( 1 - n \sin \frac 1 n \right )$
$\sum_{n=1}^\infty (-1)^n \left ( 1 - \cos \frac 1 n \right )$
$\sum_{n=1}^\infty (-1)^n \arctan \frac 1 {2n+1}$
$\sum_{n=1}^\infty (-1)^n \left ( \frac \pi 2 - \arctan(\log n) \right )$

OK, that's probably enough to get an idea of the problems that I'm struggling with. The question for each of these is if they are absolutely convergent. From the way that Abel's test is presented, I thought that it would be most useful if you had a series which was presented in as a product, similar to integration by parts, and you knew that one of those parts was convergent. None of these seem to have that form, however I attempted to force it into this form with the following example:

$1 - n \sin \frac 1 n=\frac{1}{n^2} \left(n^2-n^3 \sin \frac 1 n \right)$

$\frac {d} {dn} \left(n^2-n^3 \sin \frac 1 n \right) = 2n-3n^2 \sin \frac 1 n +n \sin \frac 1 n=n(2-3n\sin\frac 1 n)+n\sin\frac 1 n$

Now since $\lim_{n \to \infty} n \sin \frac 1 n = 1$, the derivative is always negative for all $n >N$ for some N. Therefore, we can use the convergence of $\sum \frac 1 {n^2}$ and Abel's test to conclude that the series $\sum (-1)^n \left ( 1 - n \sin \frac 1 n \right)$ is absolutely convergent.

Is this the correct approach to this type of problem? How would this be applicable to, say, $\sum \lim \left (n\sin\frac 1 n \right)$? Thanks in advance.

2. ## Re: Difficult series section - Abel/Dirichlet's Test

To me, none of these looks like a case for using the Abel/Dirichlet tests. I would use the (limit) comparison test for all of them.

Taking $\sum_{n=1}^\infty \log \left(n \sin \tfrac 1 n \right)$ as an example, start by looking at the function $\log\left(\tfrac{\sin x}x\right).$ If we can approximate this by a power of x, as $x\to0$, then by putting $x=1/n$ we can estimate the size of $\log \left(n \sin \tfrac 1 n \right).$

Using the first couple of terms in the power series $\sin x = x-\tfrac16x^3 + \ldots$, you get the approximation $\log\left(\tfrac{\sin x}x\right)\approx \log\bigl(1-\tfrac16x^2\bigr).$ Now use a power series approximation again, putting $s=\tfrac16x^2$ in the series for $\log(1-s)$, to get $\log\bigl(1-\tfrac16x^2\bigr)\approx -\tfrac16x^2.$

You will notice that that is a very casual, nonrigorous argument. But now that we have discovered that $\log\left(\tfrac{\sin x}x\right)\approx-\tfrac16x^2$, we can look at the ratio $\frac{\log\left(\tfrac{\sin x}x\right)}{x^2}$ and prove rigorously (by repeated applications of l'Hôpital's rule, for example), that $\lim_{x\searrow0}\frac{\log\left(\tfrac{\sin x}x\right)}{x^2} = -\tfrac16.$

Finally, you can put $x=1/n$ and deduce that $\lim_{n\to\infty}\tfrac{a_n}{b_n} = \tfrac16$, where $a_n = \left|\log \left(n \sin \tfrac 1 n \right)\right|$ and $b_n=\tfrac1{n^2}.$ Since $\sum b_n$ converges, it follows from the limit comparison test that $\sum\log\left(\tfrac{\sin x}x\right)$ converges absolutely.

3. ## Re: Difficult series section - Abel/Dirichlet's Test

Thank you so much! That makes complete sense. The book hasn't used approximations yet, so it's good that you explained how to get a rigorous proof out of it afterwards. We haven't done power series either, so I had not seen the method of substituting x=1/n. We did cover Taylor approximations, using little-o notation, so to stay within bounds of what we have done I can use that, which is wonderful.

I now see how the use of approximations can help in finding a function to use a limit test with. Very cool.

One other question - when considering absolute convergence and using the limit test, if I reference the continuity of the absolute value function, if $\textstyle b_n>0$, can I just write

$\lim_{n\to\infty} \frac{|a_n|}{b_n}=\left|\lim_{n\to\infty}\frac{a_n }{b_n}\right|$

4. ## Re: Difficult series section - Abel/Dirichlet's Test

Originally Posted by process91
One other question - when considering absolute convergence and using the limit test, if I reference the continuity of the absolute value function, if $\textstyle b_n>0$, can I just write

$\lim_{n\to\infty} \frac{|a_n|}{b_n}=\left|\lim_{n\to\infty}\frac{a_n }{b_n}\right|$
Yes. It is always true that if $\lim_{n\to\infty}x_n$ exists then $\lim_{n\to\infty}|x_n| = \Bigl|\lim_{n\to\infty}x_n\Bigr|.$

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